Problem: There is a railroad in Takahashi Kingdom. The railroad consists of N sections, numbered 1, 2, ..., N, and N+1 stations, numbered 0, 1, ..., N. Section i directly connects the stations i-1 and i. A train takes exactly A_i minutes to run through section i, regardless of direction. Each of the N sections is either single-tracked over the whole length, or double-tracked over the whole length. If B_i = 1, section i is single-tracked; if B_i = 2, section i is double-tracked. Two trains running in opposite directions can cross each other on a double-tracked section, but not on a single-tracked section. Trains can also cross each other at a station.

Snuke is creating the timetable for this railroad. In this timetable, the trains on the railroad run every K minutes, as shown in the following figure. Here, bold lines represent the positions of trains running on the railroad. (See Sample 1 for clarification.)

<image>

When creating such a timetable, find the minimum sum of the amount of time required for a train to depart station 0 and reach station N, and the amount of time required for a train to depart station N and reach station 0. It can be proved that, if there exists a timetable satisfying the conditions in this problem, this minimum sum is always an integer.

Formally, the times at which trains arrive and depart must satisfy the following:

* Each train either departs station 0 and is bound for station N, or departs station N and is bound for station 0.
* Each train takes exactly A_i minutes to run through section i. For example, if a train bound for station N departs station i-1 at time t, the train arrives at station i exactly at time t+A_i.
* Assume that a train bound for station N arrives at a station at time s, and departs the station at time t. Then, the next train bound for station N arrives at the station at time s+K, and departs the station at time t+K. Additionally, the previous train bound for station N arrives at the station at time s-K, and departs the station at time t-K. This must also be true for trains bound for station 0.
* Trains running in opposite directions must not be running on the same single-tracked section (except the stations at both ends) at the same time.

Constraints

* 1 \leq N \leq 100000
* 1 \leq K \leq 10^9
* 1 \leq A_i \leq 10^9
* A_i is an integer.
* B_i is either 1 or 2.

Input

The input is given from Standard Input in the following format:


N K
A_1 B_1
A_2 B_2
:
A_N B_N


Output

Print an integer representing the minimum sum of the amount of time required for a train to depart station 0 and reach station N, and the amount of time required for a train to depart station N and reach station 0. If it is impossible to create a timetable satisfying the conditions, print -1 instead.

Examples

Input

3 10
4 1
3 1
4 1


Output

26


Input

1 10
10 1


Output

-1


Input

6 4
1 1
1 1
1 1
1 1
1 1
1 1


Output

12


Input

20 987654321
129662684 2
162021979 1
458437539 1
319670097 2
202863355 1
112218745 1
348732033 1
323036578 1
382398703 1
55854389 1
283445191 1
151300613 1
693338042 2
191178308 2
386707193 1
204580036 1
335134457 1
122253639 1
824646518 2
902554792 2


Output

14829091348

Solution (in CPP):

Solution:
#include<cstdio>
#include<algorithm>
#define maxn 100005
#define LL long long
using namespace std;
int n,k,cnt,a[maxn],b[maxn],tree[maxn*4];
LL sum[maxn],t[maxn*2],l[maxn],r[maxn],dp[maxn];
void Pushdown(int i)
{
	if(!tree[i]) return;
	tree[i<<1]=tree[i<<1|1]=tree[i];
	tree[i]=0;
}
void Modify(int i,int l,int r,int ql,int qr,int d)
{
	if(l>r||ql>qr) return;
	if(ql<=l&&r<=qr) { tree[i]=d; return; }
	Pushdown(i);
	int mid=(l+r)>>1;
	if(ql<=mid) Modify(i<<1,l,mid,ql,qr,d);
	if(qr>mid) Modify(i<<1|1,mid+1,r,ql,qr,d);
}
int Query(int i,int l,int r,int p)
{
	if(l==r) return tree[i];
	Pushdown(i);
	int mid=(l+r)>>1;
	if(p<=mid) return Query(i<<1,l,mid,p);
	else return Query(i<<1|1,mid+1,r,p);
}
LL Query(int p)
{
	int c=Query(1,1,cnt,p);
	if(!c) return 0;
	return 1ll*dp[c]+(t[l[c]]-t[p]+k)%k;
}
int main()
{
	scanf("%d%d",&n,&k);
	for(int i=1;i<=n;i++)
	{
		scanf("%d%d",&a[i],&b[i]);
		sum[i]=sum[i-1]+a[i];
		if(b[i]==1&&2*a[i]>k) { printf("-1\n"); return 0; }
	}
	for(int i=1;i<=n;i++)
	{
		if(b[i]==1) l[i]=(-2*sum[i-1]%k+k)%k,r[i]=(-2*sum[i]%k+k)%k;
		else l[i]=0,r[i]=k-1;
		t[i*2-1]=l[i],t[i*2]=r[i];
	}
	sort(t+1,t+n*2+1);
	cnt=unique(t+1,t+n*2+1)-t-1;
	for(int i=1;i<=n;i++)
		l[i]=lower_bound(t+1,t+cnt+1,l[i])-t,r[i]=lower_bound(t+1,t+cnt+1,r[i])-t;
	for(LL i=n;i;i--)
	{
		dp[i]=Query(l[i]);
		if(l[i]>r[i]) Modify(1,1,cnt,r[i]+1,l[i]-1,i);
		else { Modify(1,1,cnt,1,l[i]-1,i); Modify(1,1,cnt,r[i]+1,cnt,i); }
	}
	LL ans=dp[1];
	for(int i=1;i<=cnt;i++) ans=min(ans,Query(i));
	printf("%lld\n",ans+sum[n]*2);
}